列空間、零解空間、及其基底

# 列空間、零解空間、及其基底 Row space, kernel, and their bases ![Creative Commons License](https://i.creativecommons.org/l/by/4.0/88x31.png) This work by Jephian Lin is licensed under a [Creative Commons Attribution 4.0 International License](http://creativecommons.org/licenses/by/4.0/). $\newcommand{\trans}{^\top} \newcommand{\adj}{^{\rm adj}} \newcommand{\cof}{^{\rm cof}} \newcommand{\inp}[2]{\left\langle#1,#2\right\rangle} \newcommand{\dunion}{\mathbin{\dot\cup}} \newcommand{\bzero}{\mathbf{0}} \newcommand{\bone}{\mathbf{1}} \newcommand{\ba}{\mathbf{a}} \newcommand{\bb}{\mathbf{b}} \newcommand{\bc}{\mathbf{c}} \newcommand{\bd}{\mathbf{d}} \newcommand{\be}{\mathbf{e}} \newcommand{\bh}{\mathbf{h}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bq}{\mathbf{q}} \newcommand{\br}{\mathbf{r}} \newcommand{\bx}{\mathbf{x}} \newcommand{\by}{\mathbf{y}} \newcommand{\bz}{\mathbf{z}} \newcommand{\bu}{\mathbf{u}} \newcommand{\bv}{\mathbf{v}} \newcommand{\bw}{\mathbf{w}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\nul}{\operatorname{null}} \newcommand{\rank}{\operatorname{rank}} %\newcommand{\ker}{\operatorname{ker}} \newcommand{\range}{\operatorname{range}} \newcommand{\Col}{\operatorname{Col}} \newcommand{\Row}{\operatorname{Row}} \newcommand{\spec}{\operatorname{spec}} \newcommand{\vspan}{\operatorname{span}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\idmap}{\operatorname{id}} \newcommand{\am}{\operatorname{am}} \newcommand{\gm}{\operatorname{gm}} \newcommand{\mult}{\operatorname{mult}} \newcommand{\iner}{\operatorname{iner}}$ ```python from lingeo import random_good_matrix, row_space_matrix, kernel_matrix ``` ## Main idea You are recommended to read the section _Four fundamental subspaces_ first, where you will find the definition of $\beta_R$, $\beta_K$, $\beta_C$, $\beta_L$. Let $A$ be a matrix. Let $\beta_R$ and $\beta_K$ be the standard bases of $\Row(A)$ and $\ker(A)$, respectively. We have known that 1. $\Row(A) = \vspan(\beta_R)$. 2. $\ker(A) = \vspan(\beta_K)$. In fact, both $\beta_R$ and $\beta_K$ are linearly independent. Therefore, it is fine that we call them the standard bases. Let $V$ be a subspace in $\mathbb{R}^n$ spanned by a finite set of vectors $S$. Then we may find a basis of $V^\perp$ as follows. 1. Write the matrix $A$ whose rows are the vectors in the basis. 3. Calculate $\beta_K$ and it is a basis of $V^\perp$. ## Side stories - basis of the orthogonal complement - basis of a hyperplane ## Experiments ##### Exercise 1 執行下方程式碼。令 $\left[\begin{array}{c|c} R & B \end{array}\right]$ 為 $\left[\begin{array}{c|c} A & I \end{array}\right]$ 最簡階梯形式矩陣。令 $S = \{ \br_1, \ldots, \br_3 \}$ 為 $A$ 的各列向量﹐ 而 $S' = \{ \br'_1, \ldots, \br'_3 \}$ 為 $R$ 的各列向量。  Run the code below. Let $\left[\begin{array}{c|c} R & B \end{array}\right]$ be the reduced echelon form of $\left[\begin{array}{c|c} A & I \end{array}\right]$. Let $S= \{ \bu_1, \ldots, \bu_3 \}$ be the rows of $A$ and $S' = \{ \br'_1, \ldots, \br'_3 \}$ the rows of $R$.  ```python ### code set_random_seed(0) print_ans = False m,n,r = 3,5,2 A = random_good_matrix(m,n,r) AI = A.augment(identity_matrix(m), subdivide=True) RB = AI.rref() B = RB[:,n:] print("[ A | I ] =") show(AI) print("[ R | B ] =") show(RB) if print_ans: for i in range(m): print( "r'%s = "%(i+1) + " + ".join("%s u%s"%(B[i,k], k+1) for k in range(m)) ) ``` $$ \left[\begin{array}{c|c} A & I \end{array}\right]= \left[\begin{array}{ccccc|ccc} 1 & -3 & 18 & 5 & -14 & 1 & 0 & 0\\ 3 & -8 & 49 & 15 &-39 & 0 & 1 & 0\\ -8 & 20 & -124 & -40 & 100 & 0 & 0 & 1 \end{array}\right] $$ $$ \left[\begin{array}{c|c} R & B \end{array}\right]= \left[\begin{array}{ccccc|ccc} 1 & 0 & 3 & 5 & -5 & 0 & -5 & -2\\ 0 & 1 & -5 & 0 & 3 & 0 & -2 & -\frac{3}{4}\\ 0 & 0 & 0 & 0 & 0 & 1 & -1 & -\frac{1}{4} \end{array}\right] $$ :::warning Record the values given by the code. ::: ##### Exercise 1(a) 將 $S'$ 中的每一個向量寫成 $S$ 的線性組合。  For each vector in $S'$, write it as a linear combination of $S$.  :::warning - [x] Explain why and how you obtain the answers. - [x] Use boldface, e.g., $\br$, for vectors. - [x] Since $\left[\begin{array}{c|c} R & B \end{array}\right]$ is the reduced echelon form of $\left[\begin{array}{c|c} A & I \end{array}\right]$, every row of the former matrix is a linear combination of the rows of the latter matrix. By observing the last three entries of $\left[\begin{array}{c|c} R & B \end{array}\right]$, we know that ${\bf r}_1' = 0{\bf r}_1 - 5{\bf r}_2 - 2{\bf r}_3$. Similarly, we have ... ::: ##### Answer for 1(a): Since $\left[\begin{array}{c|c} R & B \end{array}\right]$ is the reduced echelon form of $\left[\begin{array}{c|c} A & I \end{array}\right]$, every row of the former matrix is a linear combination of the rows of the latter matrix. By observing the last three entries of $\left[\begin{array}{c|c} R & B \end{array}\right]$, we know that ${\bf r}_1' = 0{\bf r}_1 - 5{\bf r}_2 - 2{\bf r}_3$. Similarly, we have ${\bf r}_2' = 0{\bf r}_1 - 2{\bf r}_2 - \frac{3}{4}{\bf r}_3$ and ${\bf r}_3' = 1{\bf r}_1 - 1{\bf r}_2 - \frac{1}{4}{\bf r}_3.$ *** ##### Exercise 1(b) 令 $B$ 的第一列為 $(b_1, \ldots, b_3)$。說明如果要把 $\left[\begin{array}{c|c} R & B \end{array}\right]$ 的第一列寫成 $\left[\begin{array}{c|c} A & I \end{array}\right]$ 各列的線性組合﹐ 使用係數 $b_1, \ldots, b_3$ 是唯一的辦法。更一般來說﹐說明 $BA = R$。（提示：令 $\left[\begin{array}{c|c} R & B \end{array}\right]$ 的各列為 $\bw'_1, \ldots, \bw'_3$、而 $\left[\begin{array}{c|c} A & I \end{array}\right]$ 的各列為 $\bw_1, \ldots, \bw_3$。解 $x_1\bw_1 + x_2\bw_2 + x_3\bw_3 = \bw'_1$。只要觀察這個方程式每個向量的最後幾項就好。）  Let $(b_1, \ldots, b_3)$ be the first row of $B$. Explain the only way to write the first row of $\left[\begin{array}{c|c} R & B \end{array}\right]$ as a linear combination of rows of $\left[\begin{array}{c|c} A & I \end{array}\right]$ is by the coefficients $b_1, \ldots, b_3$. More generally, explain why $BA = R$. Hint: Let $\bw'_1, \ldots, \bw'_3$ be the rows of $\left[\begin{array}{c|c} R & B \end{array}\right]$ and $\bw_1, \ldots, \bw_3$ the rows of $\left[\begin{array}{c|c} A & I \end{array}\right]$. Try to solve $x_1\bw_1 + x_2\bw_2 + x_3\bw_3 = \bw'_1$ for $x_1, x_2, x_3$. Note that you only need to focus on the last few entries.  :::warning :warning: Not yet done. Still need to explain $BA = R$. Done now. ::: ##### Answer 1(b): Let $\bw'_1, \ldots, \bw'_3$ be the rows of $\left[\begin{array}{c|c} R & B \end{array}\right]$ and $\bw_1, \ldots, \bw_3$ the rows of $\left[\begin{array}{c|c} A & I \end{array}\right]$. $$ \bw_1= \begin{pmatrix} 1 & -3 & 18 & 5 & -14 & 1 & 0 & 0 \end{pmatrix}, $$ $$ \bw_2= \begin{pmatrix} 3 & -8 & 49 & 15 & -39 & 0 & 1 & 0 \end{pmatrix}, $$ $$ \bw_3= \begin{pmatrix} -8 & 20 & -124 & -40 & 100 & 0 & 0 & 1 \end{pmatrix}, $$ $$ \bw'_1= \begin{pmatrix} 1 & 0 & 3 & 5 & -5 & 0 & -5 & -2 \end{pmatrix}. $$ Sovle $x_1\bw_1 + x_2\bw_2 + x_3\bw_3 = \bw'_1$ for $x_1, x_2, x_3$. By observing the last three columns, it is clear that $x_1=0,x_2=-5,x_3=-2.$ Thus, we know $$ \begin{bmatrix} 0 & -5 & 2 \end{bmatrix} \left[\begin{array}{c|c} A & I \end{array}\right] $$ is the first row of $\left[\begin{array}{c|c} R & B \end{array}\right]$. Similarly, $$ \begin{bmatrix} 0 & -2 & -\frac{3}{4} \end{bmatrix} \left[\begin{array}{c|c} A & I \end{array}\right] \text{ and } \begin{bmatrix} 1 & -1 & -\frac{1}{4} \end{bmatrix} \left[\begin{array}{c|c} A & I \end{array}\right] $$ are the second and the third rows of $\left[\begin{array}{c|c} R & B \end{array}\right]$, respectively. Therefore, $B \left[\begin{array}{c|c} A & I \end{array}\right] = \left[\begin{array}{c|c} R & B \end{array}\right]$. In other word, $BA=R.$ *** ##### Exercise 1(c) 若 $A$ 是一個 $n\times n$ 矩陣。而 $\left[\begin{array}{c|c} A & I_n \end{array}\right]$ 的最簡階梯形式矩陣是 $\left[\begin{array}{c|c} I_n & B \end{array}\right]$。藉由上一小題的結果（再次）證明若 $B$ 是 $n\times n$ 矩陣使得 $AB = I_n$﹐ 則 $BA = I_n$。  Let $A$ be an $n\times n$ matrix. Suppose $\left[\begin{array}{c|c} I_n & B \end{array}\right]$ is the reduced echelon form of $\left[\begin{array}{c|c} A & I_n \end{array}\right]$. Use the observation from the previous problem prove (again) that if $B$ is an $n\times n$ matrix such that $AB = I_n$, then $BA = I_n$.  :::warning :warning: Not yet done. Need to explain why $AB = I$ implies $\left[\begin{array}{c|c} I & B \end{array}\right]$ is the reduced echelon form of $\left[\begin{array}{c|c} A & I \end{array}\right]$. ::: **Solution:** From Exercise 1(b), we know that $\left[\begin{array}{c|c} A & B \end{array}\right]$ after reduced row echelon form$\implies$ $\left[\begin{array}{c|c} R & B \end{array}\right]$, then $BA=R.$ So $\left[\begin{array}{c|c} A & I_n \end{array}\right]$ after reduced row echelon form, we get $\left[\begin{array}{c|c} I_n & B \end{array}\right].$ Therefore $BA=I_n.$ *** ## Exercises ##### Exercise 2 執行以下程式碼。其中 $R$ 是 $A$ 的最簡階梯形式矩陣。  Run the code below. Let $R$ be the reduced echelon form of $A$.  ```python ### code set_random_seed(0) print_ans = False m,n,r = 4,5,2 A = random_good_matrix(m,n,r) R = A.rref() print("A =") show(A) print("R =") show(R) if print_ans: print("A basis of the row space can be the set of rows of") show(row_space_matrix(A)) print("A basis of the kernel can be the set of columns of") show(kernel_matrix(A)) ``` :::warning - [x] Use the following template to record the values. By running the code with `seed = ?`, we get ... ::: By running the code with `seed = 0`, we get $$A =\begin{bmatrix} 1 & -3 & 18 & 5 & -14 \\ 3 & -8 & 49 & 15 & -39 \\ -5 & 12 & -75 & -25 & 61\\2 & -4 & 26 & 10 &-22 \end{bmatrix}$$ and $$R = \begin{bmatrix} 1 & 0 & 3 & 5 & -5 \\ 0 & 1 & -5 & 0 & 3 \\ 0 & 0 & 0 & 0 & 0\\0 & 0 & 0 & 0 & 0 \end{bmatrix}. $$ *** ##### Exercise 2(a) 求出 $\Row(A)$ 的一組基底。  Find a basis of $\Row(A)$.  :::warning - [x] which --> where - [x] Add a sentence: Recall that $\beta_R$ is the set of nonzero rows in $R$. - [x] $\beta_R = \{\br_1$, $\br_2\}$ --> $\beta_R = \{\br_1, \br_2\}$ (see source code) - [x] Before the last sentence, add: Also, the nonzero rows in $R$ are independent since it is in echelon form. ::: ***Solution:*** Let $A = \begin{bmatrix} - & {\bf a}_1 & - \\ ~ & \vdots & ~ \\ - & {\bf a}_m & - \\ \end{bmatrix}$, and $R = \begin{bmatrix} - & {\bf r}_1 & - \\ ~ & \vdots & ~ \\ - & {\bf r}_m & - \\ \end{bmatrix}$. We know that $\Row(A) = \vspan(\beta_R)$, where $\beta_R$ is the standard basis of $\Row(A)$. Recall that $\beta_R$ is the set of nonzero rows in $R$. Thus, $\beta_R = \{\br_1, \br_2\}$. Since ${\bf a}_1= {\bf r}_1-3{\bf r}_2$, ${\bf a}_2= 3{\bf r}_1-8{\bf r}_2$, ${\bf a}_3= -5{\bf r}_1+12{\bf r}_2$, ${\bf a}_4= 2{\bf r}_1-4{\bf r}_2$. Also, the nonzero rows in $R$ are independent since it is in echelon form. Therefore, $\beta_R$ is a basis of $\Row(A)$. The basis of $\Row(A)$ is $$ \{ \begin{pmatrix} 1 & 0 & 3 & 5 & -5\\ \end{pmatrix}, \begin{pmatrix} 0 & 1 & -5 & 0 & 3\\ \end{pmatrix} \}. $$ *** ##### Exercise 2(b) 求出 $\ker(A)$ 的一組基底。  Find a basis of $\ker(A)$.  :::warning :warning: Your asnwer is a vector in $\ker(A)$ but not a basis of $\ker(A)$. - [x] Suppose $$ \bx = \begin{bmatrix} a\\ b\\ c\\ d\\ e\\ \end{bmatrix}, a,b,c,d,e\in\mathbb{R}. $$ is a vector such that $A\bx = \bzero$. <-- boldface $\bzero$ - [x] Since ..., ~~then~~ ... . - [x] Period in the end. ::: **Solution:** Find $\ker(A)$: Suppose $$ \bx = \begin{bmatrix} a\\ b\\ c\\ d\\ e\\ \end{bmatrix}, a,b,c,d,e\in\mathbb{R} $$ is a vector such that $A\bx=\bzero$. Since $R$ is the reduced row echelon form, $\ker(A)=\ker(R)$. Find $\bx$: By assuming $c = p$, $d = q$, $e = r$, we have $$ \bx = \begin{bmatrix} -3p-5q+5r\\ 5p-3r\\ p\\ q\\ r\\ \end{bmatrix} =p\begin{bmatrix} -3\\ 5\\ 1\\ 0\\ 0\\ \end{bmatrix} +q\begin{bmatrix} -5\\ 0\\ 0\\ 1\\ 0\\ \end{bmatrix} +r\begin{bmatrix} 5\\ -3\\ 0\\ 0\\ 1\\ \end{bmatrix}, p,q,r\in\mathbb{R}. $$ Therefore, the basis of $\ker(A)$ is $$ \left\{ \begin{bmatrix} -3\\ 5\\ 1\\ 0\\ 0\\ \end{bmatrix}, \begin{bmatrix} -5\\ 0\\ 0\\ 1\\ 0\\ \end{bmatrix},\begin{bmatrix} 5\\ -3\\ 0\\ 0\\ 1\\ \end{bmatrix} \right\}. $$ *** ##### Exercise 3 令 $$ A = \begin{bmatrix} 1 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & -1 \\ \end{bmatrix} $$ 且 $S = \{\br_1,\ldots,\br_3\}$ 為其各列向量。令 $V = \vspan(S)$，求 $V^\perp$ 的一組基底。  Let $$ A = \begin{bmatrix} 1 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & -1 \\ \end{bmatrix} $$ and $S = \{\br_1,\ldots,\br_3\}$ its rows. Let $V = \vspan(S)$. Find a basis of $V^\perp$.  :::warning :warning: Mathematical error. Every vector in $V^\perp$ is orthogonal to every vector in $\vspan(S)$. Is this true for your answer? Still not correct... the vector $(-1,-1,0)$ is not orthogonal to the vector $(0,-1,-1)$ in $S$. - [ ] Use the right macros: $span$ --> $\vspan$, $Row$ --> $\Row$, $ker$ --> $\ker$ - [ ] Put equal signs in math mode `$...$` - [ ] The paragraph about "Getting $\ker(A)$" need some more words, for example: By solving $$ \begin{bmatrix} 1 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & -1 \end{bmatrix} \begin{bmatrix}\ x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = \begin{bmatrix}\ 0 \\ 0 \\ 0 \\ \end{bmatrix}, $$ we know $$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = t \begin{bmatrix} -1\\ 1\\ -1\\ \end{bmatrix} $$ for arbitrary $r\in \mathbb{R}$. ==:x: Vector is not equal to a set!!!== ::: **Answer** We can get the reduced echelon form of $A$, $$ R = \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end{bmatrix}. $$ Since $V=\vspan(S)=\Row(A)$, we know that $\ker(A) \perp \Row(A)$. Therefore, $V^\perp=\ker(A)$. Getting $\ker(A)$: By solving $$ \begin{bmatrix} 1 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & -1 \end{bmatrix} \begin{bmatrix}\ x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = \begin{bmatrix}\ 0 \\ 0 \\ 0 \\ \end{bmatrix}, $$ we know $$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = t \begin{bmatrix} 1\\ -1\\ 1\\ \end{bmatrix} ,\ t\in \Bbb R . $$ Thus, we can have a basis of $V^\perp$, which is equal to a basis of $\ker(A)$ $$ \left\{ \begin{bmatrix} 1\\ -1\\ 1\\ \end{bmatrix} \right\}. $$ *** ##### Exercise 4 利用 zero forcing 的方法來說明 $\beta_R$ 是線性獨立的。  Use zero forcing to show that $\beta_R$ is linearly independent.  :::warning - [x] bases --> basis - [x] Consider the equation $c_1\bu_1+c_2\bu_2=\bzero$. By observing the first entries on the both sides, we get $c_1 = 0$. Then by observing ... ::: :::success Now it looks nice! ::: **Solution:** Since $\beta_R$ is the standard basis of $\Row(A)$, $$ \beta_R= \{ \begin{pmatrix} 1 & 0 & 3 & 5 & -5 \end{pmatrix}, \begin{pmatrix} 0 & 1 & -5 & 0 & 3 \end{pmatrix} \} $$ Let $$ \bu_1= \begin{pmatrix} 1 & 0 & 3 & 5 & -5 \end{pmatrix}, \bu_2= \begin{pmatrix} 0 & 1 & -5 & 0 & 3 \end{pmatrix}. $$ If $\beta_R$ is linearly independent, then the only coefficients $c_1,c_2 \in \mathbb{R}$ satisfying $$ c_1\bu_1+c_2\bu_2=\bzero $$ is $c_1=c_2=0$. Find $c_1$, $c_2$: Consider the equation $c_1\bu_1+c_2\bu_2=\bzero$ By observing the first entries on the both sides, we get $c_1=0$. Then by observing the second entries on the both sides, we get $c_2=0$. Thus $\beta_R$ is linearly independent. ##### Exercise 5 利用 zero forcing 的方法來說明 $\beta_K$ 是線性獨立的。  Use zero forcing to show that $\beta_K$ is linearly independent.  **Solution for Exercise 5:** In order to show that a matrix is linearly independent with the **zero forcing** method, we need to:  1. Let $S = \{ \bu_1, \ldots, \bu_k \}$ be some vectors in $\mathbb{R}^n$, and $S' = S$. 2. If for some $j = 1,\ldots, n$, every vector in $S'$ has its $j$-th entry zero except for one vector $\bu_i$ with its $j$-th entry nonzero, then remove $\bu_i$ from $S'$. 3. Repeat Step 2. If there is a way to remove all vectors, then $S$ is a linearly independent set.  As for a basis of a kernal, $\beta_K$, there would be a **Identity Matrix** within the basis matrix, reasoning as follows: In the process of the search for $\beta_K$, we will reduce the matrix $A$ to a reduced row echelon form $R$, let's take exercise 2(b) as an example: $$A =\begin{bmatrix} 1 & -3 & 18 & 5 & -14 \\ 3 & -8 & 49 & 15 & -39 \\ -5 & 12 & -75 & -25 & 61\\ 2 & -4 & 26 & 10 &-22 \\ \end{bmatrix}$$ $$R =\begin{bmatrix} 1 & 0 & 3 & 5 & -5 \\ 0 & 1 & -5 & 0 & 3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix} $$ :::warning - [x] Use boldface only for vectors, so change the variables into normal font type. ::: We let the lead variables be x and y, and the free variables be a,b, and c . The kernel would then be: $$\ker(A) =\begin{bmatrix} x\\ y\\ a\\ b\\ c\\ \end{bmatrix} $$ And since: $$\begin{bmatrix} 1 & 0 & 3 & 5 & -5 \\ 0 & 1 & -5 & 0 & 3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix} \begin{bmatrix} x \\ y \\ a \\ b \\ c \\ \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix}$$ We can know that: $$\begin{bmatrix} x \\ y \\ a \\ b \\ c \\ \end{bmatrix}= \begin{bmatrix} -3a-5b+5c\\ 5a-3c\\ a\\ b\\ c\\ \end{bmatrix}$$ And in order to get $\beta_K$, we let each of the free variables a, b, and c to be 1 for each entry, the result would then be: $$ \beta_K = \begin{bmatrix} -3 & -5 & 5 \\ 5 & 0 & -3 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$$ And that's the reason why every basis for a kernel would have an **Identity Matrix** within it, and with the presence of an indentity matrix, we can reduce the matrix further to facilitate the **Zero Forcing** procedure. :::warning - [x] can be reduced to zeros, which leaves us --> can be replaced by any numbers, which leaves us $$\begin{bmatrix} ? & ? & ? \\ ? & ? & ? \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$ ::: That is, in the presence of an identiy matrix, the rows denoting the leading variables, (i.e. $$ \begin{bmatrix} -3 & -5 & 5 \end{bmatrix} \text{ and } \begin{bmatrix} 5 & 0 &-3 \\ \end{bmatrix}$$) regardless of their values, can be replaced by any numbers, which leaves us: $$\begin{bmatrix} ? & ? & ? \\ ? & ? & ? \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$ By following the aforementioned **Zero Forcing Method**, we remove the 3rd, 4th and 5th rows since they have the none zero entries among the zeros in their columns. And the end result leaves us no more vectors, which is the proof that $\beta_K$ will always be linearly independent. ##### Exercise 6 令 $A$ 為一 $m \times n$ 矩陣。若 $E$ 是一個 $m\times m$ 可逆矩陣且 $B = EA$。  Let $A$ be an $m\times n$ matrix. Suppose $E$ is an $m\times m$ invertible matrix and $B = EA$.  ##### Exercise 6(a) 證明 $\Row(A) = \Row(B)$。（提示：可以把 $E$ 拆成基本矩陣當做列運算。另一個方法是說明 $B$ 的每一列都可以 $A$ 的各列的線性組合，而且反之亦然。）  Show that $\Row(A) = \Row(B)$. Hint: You may write $E$ as the product of some elementary matrices. Alternatively, you may show that each row of $B$ is a linear combination of $A$, and vice versa.  ##### Exercise 6(b) 證明以下敘述等價： 1. $A$ 的各列集合線性獨立。 2. $B$ 的各列集合線性獨立。（提示：證明 $\Col(A)^\perp = \{\bzero\}$ 和 $\Col(B)^\perp = \{\bzero\}$ 等價。）  Show that the following are equivalent: 1. The rows of $A$ form a linearly independent set. 2. The rows of $B$ form a linearly independent set. Hint: You may show that $\Col(A)^\perp = \{\bzero\}$ and $\Col(B)^\perp = \{\bzero\}$ are equivalent.  ##### Exercise 6(c) 總結來說，證明若 $A$ 的各列集合是 $\Row(A)$ 的一組基底，則 $B$ 的各列集合也是 $\Row(A)$ 的一組基底。  In summary, if the rows of $A$ form a basis of $\Row(A)$, then the rows of $B$ also form a basis of $\Row(A)$.  :::info collaboration: 1 4 problems: 4 done extra: 0.5 moderator: 0.5 quality control: 1 :::

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